EXAMPLE 1 Identify polynomial functions Decide whether the

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EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. a. h (x) = x 4 – 1 x 2 + 3 4 SOLUTION a. The function is a polynomial function that is already written in standard form. It has degree 4 (quartic) and a leading coefficient of 1.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. b. g (x) = 7 x – 3 + πx 2 SOLUTION b. The function is a polynomial function written as g(x) = πx 2 + 7 x – 3 in standard form. It has degree 2(quadratic) and a leading coefficient of π.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. c. f (x) = 5 x 2 + 3 x -1 – x SOLUTION c. The function is not a polynomial function because the term 3 x – 1 has an exponent that is not a whole number.

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If

EXAMPLE 1 Identify polynomial functions Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. d. k (x) = x + 2 x – 0. 6 x 5 SOLUTION d. The function is not a polynomial function because the term 2 x does not have a variable base and an exponent that is a whole number.

EXAMPLE 2 Evaluate by direct substitution Use direct substitution to evaluate f (x) =

EXAMPLE 2 Evaluate by direct substitution Use direct substitution to evaluate f (x) = 2 x 4 – 5 x 3 – 4 x + 8 when x = 3. f (x) = 2 x 4 – 5 x 3 – 4 x + 8 Write original function. f (3) = 2(3)4 – 5(3)3 – 4(3) + 8 Substitute 3 for x. = 162 – 135 – 12 + 8 Evaluate powers and multiply. = 23 Simplify

GUIDED PRACTICE for Examples 1 and 2 Decide whether the function is a polynomial

GUIDED PRACTICE for Examples 1 and 2 Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient. 1. f (x) = 13 – 2 x SOLUTION f (x) = – 2 x + 13 It is a polynomial function. Standard form: – 2 x + 13 Degree: 1 Type: linear Leading coefficient of – 2.

GUIDED PRACTICE 2. for Examples 1 and 2 p (x) = 9 x 4

GUIDED PRACTICE 2. for Examples 1 and 2 p (x) = 9 x 4 – 5 x – 2 + 4 SOLUTION p (x) = 9 x 4 – 5 x – 2 + 4 The function is not a polynomial function.

GUIDED PRACTICE 3. for Examples 1 and 2 h (x) = 6 x 2

GUIDED PRACTICE 3. for Examples 1 and 2 h (x) = 6 x 2 + π – 3 x SOLUTION h (x) = 6 x 2 – 3 x + π The function is a polynomial function that is already written in standard form will be 6 x 2– 3 x + π. It has degree 2 (linear) and a leading coefficient of 6. It is a polynomial function. Standard form: 6 x 2– 3 x + π Degree: 2 Type: quadratic Leading coefficient of 6

GUIDED PRACTICE for Examples 1 and 2 Use direct substitution to evaluate the polynomial

GUIDED PRACTICE for Examples 1 and 2 Use direct substitution to evaluate the polynomial function for the given value of x. 4. f (x) = x 4 + 2 x 3 + 3 x 2 – 7; x = – 2 SOLUTION f (x) = x 4 + 2 x 3 + 3 x 2 – 7; x = – 2 Write original function. f (– 2) = (– 2) 4 + 2 (– 2) 3 + 3 (– 2 )2 – 7 Substitute– 2 for x. = 16 – 16 + 12 – 7 =5 Evaluate powers and multiply. Simplify

GUIDED PRACTICE 5. for Examples 1 and 2 g (x) = x 3 –

GUIDED PRACTICE 5. for Examples 1 and 2 g (x) = x 3 – 5 x 2 + 6 x + 1; x = 4 SOLUTION g (x) = x 3 – 5 x 2 + 6 x + 1; x = 4 Write original function. g (x) = 4 3 – 5(4)2 + 6(4) + 1 Substitute 4 for x. = 64 – 80 + 24 + 1 =9 Evaluate powers and multiply. Simplify